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Suppose that you read through this year’s issues of the

Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye ISBN: 9780321629111 32

Solution for problem 21E Chapter 3

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Probability and Statistics for Engineers and the Scientists | 9th Edition | ISBN: 9780321629111 | Authors: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 21E

Suppose that you read through this year’s issues of the New York Times and record each number that appears in a news article—the income of a CEO, the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year, the age of a celebrity, and so on. Now focus on the leading digit of each number, which could be 1, 2, . . . , 8, or 9. Your first thought might be that the leading digit X of a randomly selected number would be equally likely to be one of the nine possibilities (a discrete uniform distribution). However, much empirical evidence as well as some theoretical arguments suggest an alternative probability distribution called Benford’s law: a. ?Without computing individual probabilities from this formula, show that it specifies a legitimate pmf. b. ?Now compute the individual probabilities and compare to the corresponding discrete uniform distribution. c.? ?Obtain the cdf of X. d.? ?Using the cdf, what is the probability that the leading digit is at most 3? At least 5? [Note: Benford’s law is the basis for some auditing procedures used to detect fraud in financial reporting—for example, by the Internal Revenue Service.]

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Problem 21E Answer: Step1: We have Suppose that you read through this year’s issues of the New York Times and record each number that appears in a news article—the income of a CEO, the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year, the age of a celebrity, and so on. Now focus on the leading digit of each number, which could be 1, 2, . . . , 8, or 9. Your first thought might be that the leading digit X of a randomly selected number would be equally likely to be one of the nine possibilities (a discrete uniform distribution). However, much empirical evidence as well as some theoretical arguments suggest an alternative probability distribution called Benford’s law: P(x) = P(1st digit is x) = log ( x+1), x = 1,2,3,...,9. 10 x We need to find, a. Without computing individual probabilities from this formula, show that it specifies a legitimate pmf. b. Now compute the individual probabilities and compare to the corresponding discrete uniform distribution. c. Obtain the cdf of X. d. Using the cdf, what is the probability that the leading digit is at most 3 At least 5 Step2: a). We have Benford’s law: P(x) = P(1st digit is x) = log ( x+1), x = 1,2,3,...,9. 10 x If the sum of the all probabilities is unity then we say that the above function is legitimate. Consider, P(x) = log ( x+1), x = 1,2,3,...,9 10 x Apply summation on both sides to above equation 9 9 x+1 P(x) = [log ( 10 x )] x=1 x=1 1+1 2+1 3+1 4+1 5+1 6+1 = [log (10 1 )+log 10 2 )+log (10 3 )+log 10 4 )+log (10 5 )+log (10 6 )+ log ( 7+1 )+ log ( 8+1 )+log ( 9+1 )] 10 7 10 8 10 9 = [log ( )+log ( ) + log ( )+log ( )+log ( )+log ( )+log ( )+ 7 8 10 1 10 2 10 3 10 4 10 5 10 6 10 7 log ( )+log ( 10)] 10 8 10 9 Take log common from the above equation we get 2 3 4 5 6 7 8 9 10 = log 10 ×1 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 )] = log (10) [by using multiplication rule of logarithm (log (ab) = log(a) + log(b))] 10 = 1 [loga(a) = 1] 9 Therefore, P(x) =1. x=1 The sum of the all probabilities is unity hence we can say that the above function is legitimate. The probability mass function(PMF) is given by b). The cumulative distribution function(CDF) is given by c). 1).P(X 3) = 0.6020 2).P(X5) = 1 - P(X < 5) = 1 - 0.6989 = 0.3011.

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Chapter 3, Problem 21E is Solved
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Textbook: Probability and Statistics for Engineers and the Scientists
Edition: 9
Author: Ronald E. Walpole; Raymond H. Myers; Sharon L. Myers; Keying E. Ye
ISBN: 9780321629111

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